Question

On the $x$ -axis and a dsitance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{Ax}{{\left(x^2+a^2\right)}^{3/2}}$ in the x-direction. The magnitude of gravitational potential on the x-axis at a distance $x$, taking its value to be zero at infinity, is :

• A. $\frac{A}{{\left(x^2+a^2\right)}^{1/2}}$
• B. $\frac{A}{{\left(x^2+a^2\right)}^{3/2}}$
• C. $A{\left(x^2+a^2\right)}^{3/2}$
• D. $A{\left(x^2+a^2\right)}^{1/2}$

Answer 1

Answer: (A)

Given $E_G=\frac{Ax}{{\left(x^2+a^2\right)}^{3/2}},V_{\infty }=0$
$\underset{V_{\infty }}{\overset{v_x}{\int }}dV=-\underset{\infty }{\overset{x}{\int }}{\vec{E}}_G\cdot {\vec{d}}_x$
$V_x-V_{\infty }=-\underset{\infty }{\overset{x}{\int }}\frac{Ax}{{\left(x^2+a^2\right)}^{3/2}}dx$
put $x^2+a^2=z$
$2xdx=dz$
$V_x-0=-\underset{\infty }{\overset{x}{\int }}\frac{Adz}{2(z{)}^{3/2}}={\left[\frac{A}{z^{1/2}}\right]}_{\infty }^x={\left[\frac{A}{{\left(x^2+a^2\right)}^{1/2}}\right]}_{\infty }^x$
$V_x=\frac{A}{{\left(x^2+a^2\right)}^{1/2}}-0=\frac{A}{{\left(x^2+a^2\right)}^{1/2}}$